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I am going through Quantum information approach to the Ising model: Entanglement in chains of qubits by Stelmachovic et al. In Section A.4, the authors determines the eigenvalues and eigenstates of the Hamiltonian of a spin chain on a ring expressed using the momentum representation. The Hamiltonian is as follows.

$$ H = \sum_q \left[-\lambda \cos q \left(\eta^\dagger_q \eta^\dagger_{2 \pi -q}+ 2 \eta^\dagger_q \eta_q - \eta_q \eta_{2 \pi -q}\right) - \lambda i \sin q \left(\eta^\dagger_q \eta^\dagger_{2 \pi - q} + \eta_q \eta_{2 \pi -q}\right) + 2 \left(\eta^\dagger_q \eta_q - \frac{1}{2}\right) \right] $$

For a specific value of $q$ when $q \ne 0, \pi$:

$$ H_q = -\lambda \cos q \left(2 \eta^\dagger_q \eta_q + 2 \eta^\dagger_{2 \pi -q} \eta_{2 \pi -q}\right) -\lambda i \sin q \left(2 \eta^\dagger_q \eta^\dagger_{q} + 2 \eta^\dagger_{2 \pi -q} \eta_{2 \pi -q}\right) + 2 \left(\eta^\dagger_q \eta_q + \eta^\dagger_{2\pi -q} \eta_{2\pi -q} - 1\right) $$

Now, for $H_q$, the authors define the following basis.

$$ |\phi_1\rangle = \eta^\dagger_q |0\rangle\\ |\phi_2\rangle = \eta^\dagger_{2 \pi - q} |0\rangle\\ |\phi_3\rangle = |0\rangle\\ |\phi_4\rangle = \eta^\dagger_{2\pi - q} \eta^\dagger_q |0\rangle $$

My question: What are the intuitions behind these choices?

My effort: I understand that $|\phi_3\rangle = |0\rangle$ is arbitrary. Then $\eta^\dagger_q$ and $\eta^\dagger_{2 \pi - q}$ take $|\phi_3\rangle$ to $|\phi_1\rangle$ and $|\phi_2\rangle$ respectively. Finally, we consider a basis $|\phi_4\rangle$ which is the eigenvector of the combination of $\eta^\dagger_q$ and $\eta^\dagger_{2 \pi - q}$. I understand up to this.

We didn't consider the operators $\eta^\dagger_q \eta_q$, $\eta^\dagger_{2 \pi -q} \eta_{2 \pi -q}$ because $\eta^\dagger_q \eta_q |\phi_3\rangle = \eta^\dagger_{2 \pi -q} \eta_{2 \pi -q} |\phi_3\rangle = 0$. The only operator left in $H_q$ is $\eta^\dagger_q \eta^\dagger_{q}$.

I would like to know why $\eta^\dagger_q \eta^\dagger_{q}$ was not considered while defining $|\phi_4\rangle$'s.

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Remember that $\eta^{q}$ are Fermionic operators, therefore $\eta^{2}_{q}=0=\eta^{\dagger 2}_{q}$.

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    $\begingroup$ In that case $\eta_{q}^{\dagger}\eta_{-q}^{\dagger}=-\eta_{-q}^{\dagger}\eta_{q}^{\dagger}$. As the are anti-commutating. $\endgroup$ – AMS Jun 23 '16 at 5:42
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    $\begingroup$ In that case $|0\rangle$ will be mapped to the first excited state. $\endgroup$ – AMS Jun 23 '16 at 5:51
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    $\begingroup$ $\eta^{\dagger}_{q}\eta^{\dagger}_{-q} |0\rangle = |q,-q\rangle$. $\endgroup$ – AMS Jun 23 '16 at 5:55
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    $\begingroup$ $|0\rangle$ is the ground state. It is not necessary to be single qubit state. $\endgroup$ – AMS Jun 23 '16 at 6:01
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    $\begingroup$ No, i guess. Say $\eta^{\dagger}_{q}\eta^{\dagger}_{-q} |1\rangle = |a,-a\rangle$. Hitting by $\eta^{\dagger}_{q}$ or $\eta^{\dagger}_{-q}$ from left gives zero, as annihilation (creation) operators are anti-commutating. $\endgroup$ – AMS Jun 23 '16 at 6:27

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